The correct option for the above reaction is

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(A) In the first step, acetone $(CH_3COCH_3)$ reacts with $Br_2$ in an acidic medium $(AcOH)$. Acidic halogenation of ketones typically results in mon

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$X = CH_3COCH_2Br, Y = CHBr_3, Z = CH_3CO_2Na$

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(A) In the first step, acetone $(CH_3COCH_3)$ reacts with $Br_2$ in an acidic medium $(AcOH)$. Acidic halogenation of ketones typically results in monohalogenation at the $\alpha$-carbon, yielding $X = CH_3COCH_2Br$.In the second step, the monobromoacetone $(CH_3COCH_2Br)$ reacts with $Br_2$ in a basic medium $(NaOH)$. This is a haloform reaction. The base promotes further halogenation of the $\alpha$-carbon followed by cleavage to form bromoform $(CHBr_3)$ and the corresponding carboxylate salt, which is sodium acetate $(CH_3CO_2Na)$.

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